Given
regular hexagon inscribed in a circle whose radius is 10 cm
Answer
Let a regular hexagon ABCDEF is inscribed in a circle of radius 10cm and
center O. Join O to A and O to B
In triangle OAB angle AOB=360°/6=60°.
and OA=OB= r=10cm (given) thus , angle OAB=angle OBA=x°(let)
angle OAB+angle OBA+angle AOB=180°
or x°+x°+60°=180°
or 2x°=180°-60°=120°
or. x=120°/2=60° , or angle OAB=angle OBA=angle AOB=60°.
thus ,triangle OAB is an equilateral triangle.
AB=OA =OB =10cm
